Killing Fields and Conserved Quantities along Geodesics

How symmetries of a metric yield first integrals of the geodesic equations.

This is a short note just to recall the calculation of an often useful fact in computing geodesics. First, some notation: let $(M,g)$ be a Riemannian / Lorentzian manifold, with a one-parameter group of isometries $u\mapsto(\Phi_u\colon M\to M)$. The derivative of this 1 parameter family determines a vector field $K(p)=\frac{d}{du}\big|_{u=0}\Phi_u(p)$, known as the Killing Field corresponding to these isometries. These fields provide a direct link between the symmetries of $M$ and conserved quantities¹ along its geodesics. Precisely,

To show this is constant, we wish to show the derivative with respect to the geodesic parameter $t$ vanishes. Differentiating via the product rule,

$$\frac{d}{dt}g(\dot{\gamma},K)=g\left(\frac{D}{dt}\dot{\gamma},K\right)+g\left(\dot{\gamma},\frac{D}{dt}K\right)$$

Where $D/dt$ is covariant differentiation pulled back to the real line. As $\gamma$ is a geodesic, $\frac{D}{dt}\dot{\gamma}=0$ by definition. And, as $\frac{D}{dt}$ coincides with $\nabla_{\dot{\gamma}}$ for vector fields along a curve which are induced by smooth vector fields on $M$, this becomes $\frac{d}{dt}g(\dot{\gamma},K)= 0+g(\dot{\gamma},\nabla_{\dot{\gamma}}K)$. Thus, it suffices to show

$g\left(\dot{\gamma},\nabla_{\dot{\gamma}}K\right)=0$

Here’s the big picture. The Lie derivative of the metric along a killing field is zero (we prove this in an appendix below): $\mathcal{L}_K g \equiv 0$

This directly implies via a calculation (which we also do in an appendix below) that if $X,Y$ are arbitrary vector fields on $M$ and $K$ is a killing field, $g\left(\nabla_X K,Y\right)+g\left(X,\nabla_Y K\right)=0$

Now set² $X=Y=\dot{\gamma}$ and use the symmetry of the metric tensor: $0=g\left(\nabla_{\dot{\gamma}}K,\dot{\gamma}\right)+g\left(\dot{\gamma},\nabla_{\dot{\gamma}}K\right)=2g(\dot{\gamma},\nabla_{\dot{\gamma}}K)$

That’s it! $g(\dot{\gamma},\nabla_{\dot{\gamma}}K)$ vanishes, and so $g(\dot{\gamma},K)$ is constant along geodesics as claimed.

Appendix: Lie Derivative of the Metric

Here we prove the fact that underlies our whole calculation:

The intuition for this result is clear from the definitions involved:

• The Lie derivative of an object measures how it changes when flowing along a vector field
• Flowing along a Killing Field generates a 1-parameter group of isometries
• Isometries do not change the metric
• Thus, you should expect to notice no changes in the metric tensor, the Lie derivative should be zero.

Here we turn this intuition into a short proof, using the geometric, or coordinate-invariant description of the Lie derivative. Setting up notation, let $K$ be a killing field and $\Phi_t$ the associated flow by isometries. The Lie derivative of $g$ at a point $p$ is calculated by comparing $g_p$ to $g$ a bit further along the flow $g_{\Phi_t(p)}$. As these two tensors are based at different points of $M$ we cannot compare them directly, but instead must pull back to p. That is, the well-defined difference is

$\left[(\Phi_t)^\ast g_{\Phi_t(p)}\right]_p - g_p$. This difference tends to zero as $t\to 0$, and the derivative is defined as usual, as the rate this quantity vanishes in comparison to $t$:

$(\mathcal{L}_kg)_p = \lim_{t\to 0}\frac{\left[(\Phi_t)^\ast g_{\Phi_t(p)}\right] - g_p}{t}$

Alright, with the formal definition out of the way, we can turn to the case of interest, where $K$ is a killing field so $\Phi_t$ is a flow by isometries. By the very definition of isometry³ being a map which leaves the Riemannian metric invariant, we have $(\Phi_t)^\ast g_{\Phi_t(p)}=g_p$, so the numerator of our difference quotient is constant, and equal to zero. Thus, the limit is zero, and the Lie derivative of $g$ is zero, as claimed.

Appendix: Killing Fields and the Metric

Here we prove that if $K$ is a killing field and $X,Y$ are arbitrary vector fields on $M$ then $g(\nabla_X K,Y)+g(X,\nabla_Y K)=0$. In fact, we give a computation of the metric’s Lie derivative along an arbitrary vector field, and then specify to the required case using that this quantity vanishes along Killing fields.

First, a note on remembering how operators (like the Lie derivative, or covariant derivative) extend to general tensors. Say that $T$ is some tensor that takes in vector fields $X,Y$ and $\star$ is some operator. If you think of applying the tensor $T$ as a kind of multiplication (for instance, when $T$ is written like a generalized matrix), then the Leibniz rule for differentiating a product implies a formula for $\star$ applied to the smooth function $T(X,Y)$:

$\star\left(T(X,Y)\right)=[\star T](X,Y)+T(\star X, Y)+T(X,\star Y)$

If we already know how to define $\star$ on smooth functions and on (co-)Vector fields, we can view such a rule as implicitly defining $\star$ on the tensor $T$: $[\star T](X,Y)=\star(T(X,Y))-T(\star X,Y)-T(X,\star Y)$

We use this to explicitly compute the derivative $\mathcal{L}_V$ of the metric tensor $g$ below.

Footnotes

1. Precisely, $g(\dot{\gamma},K)$ is shorthand for the function $\mathbb{R}\to\mathbb{R}$ given by $t\mapsto g_{\gamma(t)}(\dot{\gamma(t)},K(\gamma(t)))$. We prove this function is constant. ↩

2. Really, set $X$ and $Y$ to smooth extensions of $\dot{\gamma}$ to $M$. ↩

3. A map $f\colon M\to M$ is an isometry if it preserves the metric tensor: $g_p(X,Y)=g_{f(p)}(f_\star X, f_\star Y)$ for all $p\in M$ and $X,Y\in T_p M$.
   Recall $f^\star g$ along a map $f$ is defined as $(f^\star g)_p(X,Y)=f^\star g_{f(p)}(X,Y):=g_{f(p)}(f_\star X, f_\star Y)$. Thus, for an isometry $(f^\star g)(X,Y)=g(X,Y)$. ↩